\(\int_0^{\pi / 2} \frac{d x}{5+4 \sin x}=A \tan ^{-1} B\), then \(A+B=\)

Let \(I=\int_0^{\pi / 2} \frac{d x}{5+4 \sin x}\)
Put \(\tan \frac{\mathrm{x}}{2}=\mathrm{t} \Rightarrow \sec ^2 \frac{\mathrm{x}}{2}\left(\frac{1}{2}\right) \mathrm{dx}=\mathrm{dt}\)
\(
\therefore \mathrm{dx}=\frac{2 \mathrm{dt}}{1+\mathrm{t}^2} \text { and } \sin \mathrm{x}=\frac{2 \mathrm{t}}{1+\mathrm{t}^2}
\)
When \(\mathrm{x}=0, \mathrm{t}=0\) and when \(\mathrm{x}=\frac{\pi}{2}, \mathrm{t}=1\)
\(\therefore=\int_0^1 \frac{1}{5+4\left(\frac{2 \mathrm{t}}{1+\mathrm{t}^2}\right)} \times \frac{2 \mathrm{dt}}{1+\mathrm{t}^2}=2 \int_0^1 \frac{\mathrm{dt}}{5+5 \mathrm{t}^2+8 \mathrm{t}}=\) \(\frac{2}{5} \int_0^1 \frac{\mathrm{dt}}{\mathrm{t}^2+\frac{8}{5} \mathrm{t}+1} \)
\( =\frac{2}{5} \int_0^1 \frac{\mathrm{dt}}{\mathrm{t}^2+\frac{8}{5} \mathrm{t}+\frac{16}{25}+\frac{9}{25}}=\frac{2}{5} \int_0^1 \frac{\mathrm{dt}}{\left(\mathrm{t}+\frac{4}{5}\right)^2+\left(\frac{3}{5}\right)^2}\)
\(\therefore I=\frac{2}{5} \times \frac{1}{\left(\frac{3}{5}\right)}\left[\tan ^{-1}\left[\frac{\mathrm{t}+\frac{4}{5}}{\left(\frac{3}{5}\right)}\right]\right]_0^1=\) \(\frac{2}{3}\left[\tan ^{-1}\left(\frac{5 \mathrm{t}+4}{3}\right)\right]_0^1 \)
\( =\frac{2}{3}\left[\tan ^{-1} 3-\tan ^{-1} \frac{4}{3}\right]=\frac{2}{3} \tan ^{-1}\left[\frac{3-\left(\frac{4}{3}\right)}{1+3\left(\frac{4}{3}\right)}\right] \)
\( =\frac{2}{3} \tan ^{-1}\left(\frac{5}{3} \times \frac{1}{5}\right)=\frac{2}{3} \tan ^{-1}\left(\frac{1}{3}\right)\)
Comparing with given data, we get
\(
\mathrm{A}=\frac{2}{3}, \mathrm{~B}=\frac{1}{3} \Rightarrow \mathrm{A}+\mathrm{B}=1
\)